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March 28, 2017

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Find all solutions in the interval 0 degrees<θ<360 degrees. If rounding necessary, round to the nearest tenth of a degree. 17sec2 θ − 15tanθsecθ − 15 = 0

  • Trig - ,

    17sec^2θ - 15secθ tanθ - 15 = 0
    15secθ tanθ = 17sec^2θ - 15
    225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225
    225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225
    64sec^4θ - 285sec^2θ + 225 = 0

    That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.

    However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.

    In radians, I get .159,1.000,2.141,2.982

  • Trig - ,

    17sec2 θ − 15tanθsecθ − 15 = 0
    17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ) - 15 = 0
    times cos^2 Ø
    17 - 15sinØ - 15cos^2 Ø = 0
    17 - 15sinØ - 15(1 - sin^2 Ø) = 0
    15sin^2 Ø - 15sinØ + 2 = 0
    sinØ = (15 ± √105)/30
    sinØ = .158435 or sinØ = .841565
    Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°

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